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The Mean Value Theorem (MVT) is a fundamental concept in calculus, which bridges the gap between derivatives and the shape of a graph. It formalizes the intuitive idea that on a smooth continuous curve, there's at least one point where the tangent to the curve (the instantaneous rate of change) is equal to the curve's average rate of change over an interval.

The theorem states: if a function is continuous on a closed interval \[a, b\] and differentiable on the corresponding open interval \(a, b\), then there exists at least one point \(c\) in \(a, b\) where \(f'(c) = \frac{f(b) - f(a)}{b - a}\).

Why is this important? MVT not only helps in proving inequalities, like in our exercise, but also in understanding the behavior of functions over an interval. When applied, it requires identifying a function's continuity and differentiability—key concepts that dictate when and how MVT can be used.

Continuity and differentiability are crucial concepts in understanding calculus and applying the Mean Value Theorem. A function is said to be continuous at a point if you can draw its graph at that point without lifting your pencil. More formally, a function \(f\) is continuous at a point \(x = a\) if \(\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)\).

For a function to be differentiable at a point, the function's derivative at that point must exist. This means the function must have a defined and non-infinite rate of change at that point, resulting in a distinct tangent line.

Connection to the MVT

The MVT requirement for the theorem's application is that a function must be continuous on a closed interval and differentiable on the open interval within it. It's like saying that to apply MVT, the function must behave nicely—without any jumps, breaks, or sharp turns—over the interval we're considering.

Absolute value measures the magnitude of a real number without regard to its sign. It's denoted by \( |x| \), which is always nonnegative. One of the fundamental properties of absolute value is that \( |ab| = |a| \cdot |b| \), which means the absolute value of a product is the product of the absolute values.

The absolute value function is always continuous, but it is not differentiable at the point \(x = 0\) because there's a sharp corner in its graph. It splits the real line into two cases: For \(x \geq 0\), \( |x| = x \), and for \(x < 0\), \( |x| = -x \).

Role in Inequalities

Absolute value properties are particularly useful when dealing with inequalities, as seen in the textbook solution. By applying \( |ab| = |a| \cdot |b| \), the solution makes an inequality easier to understand. Moreover, knowing that the sine function ranges between -1 and 1, we can assert that the absolute value of the sine function, \( |\sin(x)| \), will not exceed 1, a key step in proving the exercise's statement.